Solve k^2/2k-10+25/10-2k

$$\frac { k ^ { 2 } } { 2 k - 10 } + \frac { 25 } { 10 - 2 k }$$

Evaluate

$$\frac{k+5}{2}$$

Short Solution Steps

Factor 2k-10. Factor 10-2k.

$$\frac{k^{2}}{2\left(k-5\right)}+\frac{25}{2\left(-k+5\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(k-5\right) and 2\left(-k+5\right) is 2\left(k-5\right). Multiply \frac{25}{2\left(-k+5\right)} times \frac{-1}{-1}.

$$\frac{k^{2}}{2\left(k-5\right)}+\frac{25\left(-1\right)}{2\left(k-5\right)}$$

Since \frac{k^{2}}{2\left(k-5\right)} and \frac{25\left(-1\right)}{2\left(k-5\right)} have the same denominator, add them by adding their numerators.

$$\frac{k^{2}+25\left(-1\right)}{2\left(k-5\right)}$$

Do the multiplications in k^{2}+25\left(-1\right).

$$\frac{k^{2}-25}{2\left(k-5\right)}$$

Factor the expressions that are not already factored in \frac{k^{2}-25}{2\left(k-5\right)}.

$$\frac{\left(k-5\right)\left(k+5\right)}{2\left(k-5\right)}$$

Cancel out k-5 in both numerator and denominator.

$$\frac{k+5}{2}$$

Factor

$$\frac{k+5}{2}$$